Quantum Relative Entropy, Data Processing, and Strong Subadditivity

If $d\ge 2$, Euler's identity [F5] and the integer-power formula [F3] give ${\omega }^m=\exp (2\pi im\mathrm{/}d)$ for every integer $m$. By [F4], this equals $1$ exactly when $d$ divides $m$, so $\omega$ has exact order $d$ and $\mathrm{\mid }\omega \mathrm{\mid }=1$. The shift $X$ permutes the orthonormal basis and the phase $Z$ multiplies its vectors by unit scalars, so both are unitary. For $d=1$ both are the identity. For $d\ge 2$, $d^{-1}{\sum }_{q=0}^{d-1}{\omega }^{qm}$ is $1$ when $d\mid m$; otherwise multiplication by $1-{\omega }^m\mathrm{\ne }0$ makes the geometric sum telescope to $1-{\omega }^{md}=0$, so it is $0$.

Write $E_{ab}=\mathrm{\mid }{e}_a⟩⟨e_b\mathrm{\mid }$ and expand $R={\sum }_{a,b=0}^{d-1}{R}_{ab}\otimes E_{ab}$ in the product basis. Direct action on basis vectors gives $Z^q{E}_{ab}(Z^q{)}^{†}={\omega }^{q(a-b)}{E}_{ab}$ and $X^p{E}_{ab}(X^p{)}^{†}=E_{a+p,b+p}$, with indices modulo $d$. Averaging first over $q$ using step 1.1 removes every $a\mathrm{\ne }b$ term and keeps $R_{aa}\otimes E_{aa}$.

For each remaining diagonal unit, $d^{-1}{\sum }_{p=0}^{d-1}{E}_{a+p,a+p}=I_K\mathrm{/}d$. Therefore the full average is ${\sum }_a{R}_{aa}\otimes I_K\mathrm{/}d$. By [F1], ${\sum }_a{R}_{aa}={\mathrm{Tr}}_{K}R$, giving the identity. When $d=1$, [F1] says ${\mathrm{Tr}}_{K}R$ is the unique coefficient $R_{00}$, so the same equality holds.

View ${\rho }_{ABC}$ first across the grouped bipartition $A\mathrm{\mid }BC$: [F9] makes ${\rho }_A={\mathrm{Tr}}_{BC}{\rho }_{ABC}$ and ${\rho }_{BC}={\mathrm{Tr}}_A{\rho }_{ABC}$ states. Applying [F9] across $AB\mathrm{\mid }C$ also makes ${\rho }_{AB}={\mathrm{Tr}}_C{\rho }_{ABC}$ a state. Applying it once more to ${\rho }_{AB}$ across $A\mathrm{\mid }B$ makes ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$ a state. This is also ${\mathrm{Tr}}_C{\rho }_{BC}$: in the ordered product basis [F6]–[F7], both have coefficient ${\sum }_{a,c}({\rho }_{ABC}{)}_{(a,b,c),(a,b^{\mathrm{\prime }},c)}$, since the finite sums may be reordered. The operator ${\rho }_A\otimes {\rho }_{BC}$ is also a state: spectral decompositions of the two positive factors give nonnegative product eigenvalues, and their sum is $\mathrm{tr}({\rho }_A)\mathrm{tr}({\rho }_{BC})=1$. Applying [F2] to the bipartite states ${\rho }_{A\mathrm{\mid }BC}$ and ${\rho }_{AB}$ gives support inclusion for both relative entropies $D({\rho }_{ABC}\mathrm{\parallel }{\rho }_A\otimes {\rho }_{BC})$ and $D({\rho }_{AB}\mathrm{\parallel }{\rho }_A\otimes {\rho }_B)$; [F1] then makes both finite.

Set $X={\rho }_{ABC}$ and $Y={\rho }_A\otimes {\rho }_{BC}$. By [F6] and the ordered product basis [F7], ${\mathrm{Tr}}_{C}X={\rho }_{AB}$ and ${\mathrm{Tr}}_{C}Y={\rho }_A\otimes {\mathrm{Tr}}_C{\rho }_{BC}={\rho }_A\otimes {\rho }_B$: the latter equality follows entrywise by summing the C-diagonal block of ${\rho }_{BC}$ while the A-factor matrix coefficient remains fixed. Apply [F4] to this pair to obtain $D({\rho }_{AB}\mathrm{\parallel }{\rho }_A\otimes {\rho }_B)\le D({\rho }_{ABC}\mathrm{\parallel }{\rho }_A\otimes {\rho }_{BC})\mathrm{.}$

Both sides in step 2.1 are finite by step 1.1, so [F3] identifies them as $S({\rho }_A)+S({\rho }_B)-S({\rho }_{AB})$ and $S({\rho }_A)+S({\rho }_{BC})-S({\rho }_{ABC})$, respectively. Cancel the common finite term $S({\rho }_A)$ and rearrange the inequality to obtain $S({\rho }_{AB})+S({\rho }_{BC})\ge S({\rho }_B)+S({\rho }_{ABC})$.

Let $Q_{BC}:H_B\otimes H_C\to H_C\otimes H_B$ be the two-factor swap and let $P=I_A\otimes Q_{BC}:H_A\otimes H_B\otimes H_C\to H_A\otimes H_C\otimes H_B$. Put ${\tau }_{ACB}=P{\rho }_{ABC}{P}^{†}$. By [F3], $P$ is unitary; therefore an orthonormal eigenbasis of ${\rho }_{ABC}$ is carried to an orthonormal eigenbasis of ${\tau }_{ACB}$ with the same eigenvalues, and [F5] gives $S({\tau }_{ACB})=S({\rho }_{ABC})$. The $AC$ marginal of $\tau$ is exactly ${\rho }_{AC}$ by the definition in [F1].

In the product basis, tracing out A from $\tau$ gives ${\tau }_{CB}=Q_{BC}{\rho }_{BC}{Q}_{BC}^{†}$: its $(c,b),(c^{\mathrm{\prime }},b^{\mathrm{\prime }})$ coefficient is the sum over $a$ of the $\rho$ coefficients with indices $(a,b,c),(a,b^{\mathrm{\prime }},c^{\mathrm{\prime }})$, which is precisely the swapped coefficient. Thus [F3] and the eigenvalue argument in step 1.1 give $S({\tau }_{CB})=S({\rho }_{BC})$. Tracing out both A and B gives the same C marginal ${\tau }_C={\rho }_C$, since the finite sums over their basis indices are merely reordered.

Apply [F2] to the tripartite state $\tau$ with middle register C. It yields $S({\tau }_{AC})+S({\tau }_{CB})-S({\tau }_C)-S({\tau }_{ACB})\ge 0$. Substituting the equalities in steps 1.1 and 2.1 gives exactly the expression in [F1], so $I(A:B\mid C{)}_{\rho }\ge 0$.